题解 | #数字颠倒#

//取整数的每一位可用%

#include <stdio.h>

int main()
{
    int num = 0;
    char str[32] = {0};
    int i = 0;
    if(scanf("%d", &num))
    {
        if(num > 0)
        {
            while(num)
            {
            	//取num的每一位
                str[i] = (num % 10) + '0';    //+'0'是为了把数字转换为字符
                i++;
                num /= 10;
            }
        }
        else
        {
            str[i] = num + '0';
        }
        
        printf("%s\n", str);
    }
    return 0;
}